https://session mastering Chem 321 Physical Chemistry Thermodynamics 11 nometroo

Question

https://session mastering Chem 321 Physical Chemistry Thermodynamics 11 nometrook Problem 11.12 and 90 C, the potential of the cell Pt(s) I H2(g, 100 atm) l Hci (aA.m 0.100) lAgci s) Ag(s) is Write the cell reaction Express your answers as a chomical expression. Identify all of the phases in your answer. My Answers UR Part B Calculate AGR for the cell reaction at 55 C. Express your answer with the appropriate units. AGR Value Units My Answers Calculate AS for the cea reaction at 55 C. Express your answer with the appropriate units. As Value Units Submit My Answers Part D Calculate AH for the cell reaction at 55 C. Express your answer with the appropriate units.

Explanation / Answer

Part A:

The given cell is Pt (s)H2 (g, f = 1.00 atm)HCl (aq, m = 0.100)AgCl (s)Ag (s)

The two half reactions are:

AgCl (s) + e- -----> Ag (s) + Cl- (aq); E0 = 0.22 V

2 H+ (aq) + 2 e- -----> H2 (g); E0 = 0.00 V

Since in a cell representation, the anode is written first (oxidation) followed by the cathode (reduction), the reaction is

2 AgCl (s) + H2 (g) -----> 2 Ag (s) + 2 H+ (aq) + 2 Cl- (aq)

Part B:

Plug in t = 55C and find out E as

E (V) = 0.35510 – (0.3422*10-4)*(55) - (3.2347*10-6).(55)2 + (5.314*10-9).(55)3 = 0.3425488

Find GR as GR = -nFE where n = number of moles of electrons transferred = 2 moles and F = 1 Faraday of electricity = 96485 C = 96485 J/V.mol and E = 0.3425488 V.

Plug in values:

GR = -(2 mol).(96485 J/V.mol).(0.3425488 V) = -66,101.64194 J = -66.1016 kJ -66.102 kJ (ans).

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